Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 11x}{x + 8} = \dfrac{-23x - 32}{x + 8}$
Explanation: Multiply both sides by $x + 8$ $ \dfrac{x^2 - 11x}{x + 8} (x + 8) = \dfrac{-23x - 32}{x + 8} (x + 8)$ $ x^2 - 11x = -23x - 32$ Subtract $-23x - 32$ from both sides: $ x^2 - 11x - (-23x - 32) = -23x - 32 - (-23x - 32)$ $ x^2 - 11x + 23x + 32 = 0$ $ x^2 + 12x + 32 = 0$ Factor the expression: $ (x + 4)(x + 8) = 0$ Therefore $x = -4$ or $x = -8$ However, the original expression is undefined when $x = -8$. Therefore, the only solution is $x = -4$.